3.855 \(\int \frac{1}{x^4 (a-b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{7 b^{3/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{2 a^{5/2} \sqrt [4]{a-b x^2}}-\frac{7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac{7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}+\frac{2}{a x^3 \sqrt [4]{a-b x^2}} \]

[Out]

2/(a*x^3*(a - b*x^2)^(1/4)) - (7*(a - b*x^2)^(3/4))/(3*a^2*x^3) - (7*b*(a - b*x^2)^(3/4))/(2*a^3*x) - (7*b^(3/
2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*a^(5/2)*(a - b*x^2)^(1/4))

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Rubi [A]  time = 0.0451057, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {290, 325, 229, 228} \[ -\frac{7 b^{3/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{2 a^{5/2} \sqrt [4]{a-b x^2}}-\frac{7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac{7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}+\frac{2}{a x^3 \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a - b*x^2)^(5/4)),x]

[Out]

2/(a*x^3*(a - b*x^2)^(1/4)) - (7*(a - b*x^2)^(3/4))/(3*a^2*x^3) - (7*b*(a - b*x^2)^(3/4))/(2*a^3*x) - (7*b^(3/
2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*a^(5/2)*(a - b*x^2)^(1/4))

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a-b x^2\right )^{5/4}} \, dx &=\frac{2}{a x^3 \sqrt [4]{a-b x^2}}+\frac{7 \int \frac{1}{x^4 \sqrt [4]{a-b x^2}} \, dx}{a}\\ &=\frac{2}{a x^3 \sqrt [4]{a-b x^2}}-\frac{7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}+\frac{(7 b) \int \frac{1}{x^2 \sqrt [4]{a-b x^2}} \, dx}{2 a^2}\\ &=\frac{2}{a x^3 \sqrt [4]{a-b x^2}}-\frac{7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}-\frac{7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac{\left (7 b^2\right ) \int \frac{1}{\sqrt [4]{a-b x^2}} \, dx}{4 a^3}\\ &=\frac{2}{a x^3 \sqrt [4]{a-b x^2}}-\frac{7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}-\frac{7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac{\left (7 b^2 \sqrt [4]{1-\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1-\frac{b x^2}{a}}} \, dx}{4 a^3 \sqrt [4]{a-b x^2}}\\ &=\frac{2}{a x^3 \sqrt [4]{a-b x^2}}-\frac{7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}-\frac{7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac{7 b^{3/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{2 a^{5/2} \sqrt [4]{a-b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0107757, size = 55, normalized size = 0.44 \[ -\frac{\sqrt [4]{1-\frac{b x^2}{a}} \, _2F_1\left (-\frac{3}{2},\frac{5}{4};-\frac{1}{2};\frac{b x^2}{a}\right )}{3 a x^3 \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a - b*x^2)^(5/4)),x]

[Out]

-((1 - (b*x^2)/a)^(1/4)*Hypergeometric2F1[-3/2, 5/4, -1/2, (b*x^2)/a])/(3*a*x^3*(a - b*x^2)^(1/4))

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( -b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(-b*x^2+a)^(5/4),x)

[Out]

int(1/x^4/(-b*x^2+a)^(5/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{2} + a\right )}^{\frac{5}{4}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^2 + a)^(5/4)*x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b x^{2} + a\right )}^{\frac{3}{4}}}{b^{2} x^{8} - 2 \, a b x^{6} + a^{2} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((-b*x^2 + a)^(3/4)/(b^2*x^8 - 2*a*b*x^6 + a^2*x^4), x)

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Sympy [C]  time = 1.38744, size = 34, normalized size = 0.27 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{5}{4} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{3 a^{\frac{5}{4}} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(-b*x**2+a)**(5/4),x)

[Out]

-hyper((-3/2, 5/4), (-1/2,), b*x**2*exp_polar(2*I*pi)/a)/(3*a**(5/4)*x**3)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{2} + a\right )}^{\frac{5}{4}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^2 + a)^(5/4)*x^4), x)